## $$A, B,$$ and $$C$$ are consecutive odd integers such that $$A < B < C.$$ If $$A + B + C = 81,$$ then $$A + C =$$

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### $$A, B,$$ and $$C$$ are consecutive odd integers such that $$A < B < C.$$ If $$A + B + C = 81,$$ then $$A + C =$$

by Gmat_mission » Thu Sep 24, 2020 2:33 am

00:00

A

B

C

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$$A, B,$$ and $$C$$ are consecutive odd integers such that $$A < B < C.$$ If $$A + B + C = 81,$$ then $$A + C =$$

A) 52

B) 54

C) 56

D) 58

E) 60

Source: Magoosh

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### Re: $$A, B,$$ and $$C$$ are consecutive odd integers such that $$A < B < C.$$ If $$A + B + C = 81,$$ then $$A + C =$$

by [email protected] » Thu Oct 01, 2020 9:59 am
Gmat_mission wrote:
Thu Sep 24, 2020 2:33 am
$$A, B,$$ and $$C$$ are consecutive odd integers such that $$A < B < C.$$ If $$A + B + C = 81,$$ then $$A + C =$$

A) 52

B) 54

C) 56

D) 58

E) 60

Solution:

Since A, B, and C are consecutive odd integers, which constitute an evenly-spaced set, we see that B must be the average, or 81/3 = 27. Thus, A = 25 and C = 29, and A + C = 25 + 29 = 54.